Set the purchase price to a;
(a+2 15)÷0.9 =(a- 125)÷0.8；
8(a+2 15)= 9(a- 125)；
a = 2845
So the purchase price is 2845 yuan.
Mathematical solution equation:
1. matching method (partial quadratic equation can be solved).
2. Formula method (which can solve partial quadratic equation).
3. Factorization method (which can solve partial quadratic equations).
4. Open method (which can solve all quadratic equations in one variable) The solution of quadratic equations in one variable is really bad (you can buy fx-500 calculator or Casio's 99 1 solution equation, but you need a general form).
First, the main points of knowledge:
One-dimensional quadratic equation and one-dimensional linear equation are both integral equations, which are a key content of junior high school mathematics and the basis of studying mathematics in the future, and should be paid attention to by students.
The general form of unary quadratic equation is: ax 2+bx+c = 0, (a≠0), which is an integral equation with only one unknown and the highest degree of the unknown is 2.
The basic idea of solving quadratic equations with one variable is to simplify them into two quadratic equations with one variable. Quadratic equation with one variable has four solutions.
Methods: 1, direct Kaiping method; 2. Matching method; 3. Formula method; 4. Factorial decomposition method.
Second, detailed methods and examples:
1, direct Kaiping method:
The direct Kaiping method is a method to solve a quadratic equation with a direct square root. Solving shape by direct Kaiping method.
For example, the equation of (x-m)2=n(n≥0) has a solution of x = m √ n;
Example 1. Solve the equation (1) (3x+1) 2 = 7 (2) 9x2-24x+16 =1; 1
Analysis: (1) This equation is obviously easy to do by direct flattening, (2) The left side of the equation is completely flat (3x-4) 2, and the right side =11>; 0, so this equation can also be solved by direct Kaiping method.
(1) solution: (3x+1) 2 = 7;
∴(3x+ 1)^2=7；
∴ 3x+ 1 = √ 7 (be careful not to lose the solution);
∴x=.；
The solution of the original equation is x 1=. ，x2 =。 ;
(2) Solution: 9X2-24x+16 =11;
∴(3x-4)^2= 1 1；
∴3x-4=√ 1 1；
∴x=.；
The solution of the original equation is x 1=. ，x2 =。 ;
2. Matching method: the equation AX 2+BX+C = 0 (A ≠ 0) is solved by matching method;
First, move the constant c to the right of the equation: ax 2+bx =-c;
Convert the quadratic term into1:x2+x =-;
Add half the square of the coefficient of the first term on both sides of the equation: x2+x+() 2 =-+() 2;
The left side of the equation becomes completely flat: (x+) 2 =;
When b2-4ac≥0, x+=+
∴x=. (This is the root formula);
Example 2. Solving equation 3x2-4x2 = 0 by collocation method;
Solution: Move the constant term to the right of equation 3x 2-4x = 2;
Convert the quadratic term into1:x2-x =;
Add the square of half of the coefficient of the first order term to both sides of the equation: x2-x+() 2 =+() 2;
Formula: (x-) 2 =;
Direct square: x-=+/-;
∴x=；
The solution of the original equation is x 1=, x2 =. ;
3. Formula method: the quadratic equation of one variable is transformed into the general form of AX 2+BX+C, and then the values of various coefficients A, B and C are substituted into the formula for finding the root to get the root of the equation.
When b 2-4ac > 0, the root formula is x 1 = [-b+√ (b 2-4ac)]/2a, and x2 = [-b-√ (b 2-4ac)]/2a (two unequal real roots);
When b 2-4ac = 0, the root formula is x 1=x2=-b/2a (two equal real roots);
When b 2-4ac example 3. Solve equation 2x2-8x =-5 by formula method;
Solution: Turn the equation into a general form: 2x2-8x+5 = 0;
∴a=2,b=-8,c=5；
b^2-4ac=(-8)2-4×2×5=64-40=24>； 0;
∴x===；
The solution of the original equation is x 1=, x2 =. ;
4. Factorial decomposition method: the quadratic trinomial on one side of the equation is decomposed into the product of two linear factors, so that the two linear factors are equal to zero respectively, and two linear equations are obtained. The roots obtained by solving these two linear equations are the two roots of the original equation. This method of solving a quadratic equation with one variable is called factorization.
Example 4. Solve the following equation by factorization:
( 1)(x+3)(x-6)=-8(2)2x^2+3x=0；
(3) 6x 2+5x-50 = 0 (optional research) (4) x 2-4x+4 = 0 (optional research);
(1) solution: (x+3) (x-6) =-8;
X 2-3x- 10 = 0 (the equation has a quadratic trinomial on the left and zero on the right);
(x-5)(x+2)=0 (factorization factor on the left side of the equation);
∴x-5=0 or x+2=0 (converted into two linear equations);
∴x 1=5, x2=-2 is the solution of the original equation.
(2) Solution: 2x 2+3x = 0；;
X(2x+3)=0 (factorizing the left side of the equation by raising the common factor);
∴x=0 or 2x+3=0 (converted into two linear equations);
∴x 1=0, x2=- is the solution of the original equation.
Note: Some students easily lose the solution of x=0 when doing this kind of problem. It should be remembered that there are two solutions to the quadratic equation of one variable.
(3) Solution: 6x2+5x-50 = 0;
(2x-5)(3x+ 10)=0 (special attention should be paid to symbols when factorizing factors by cross multiplication);
* 2x-5 = 0 or 3x+10 = 0;
∴x 1=, x2=- is the solution of the original equation.
(4) solution: x 2-4x+4 = 0 (∵ 4 can be decomposed into 2.2, ∴ this problem can be factorized);
(x-2)(x-2)= 0；
∴x 1=2, x2=2 is the solution of the original equation.
Summary:
Usually, factorization is the most commonly used method to solve quadratic equations with one variable. When factorization is applied, the equation is written in a general form and the quadratic coefficient is turned into a positive number. Direct leveling method is the most basic method.
Formula and collocation are the most important methods. Formula method is suitable for any quadratic equation with one variable (some people call it universal method). When using the formula method, the original equation must be transformed into a general form to determine the coefficient, and before using the formula, the value of the discriminant should be calculated to judge whether the equation has a solution.
Matching method is a tool to derive formulas. After mastering the formula method, we can directly use the formula method to solve the quadratic equation of one variable, so we generally don't need to use the matching method to solve the quadratic equation of one variable. However, collocation method is widely used in the study of other mathematical knowledge, and it is one of the three important mathematical methods required to be mastered in junior high school, so we must master it well. Three important mathematical methods: method of substitution, collocation method and undetermined coefficient method.
Example 5. Solve the following equations in an appropriate way.
( 1)4(x+2)^2-9(x-3)^2=0(2)x^2+2x-3=0；
(3)x2-2x =-(4)4x 2-4mx- 10x+m2+5m+6 = 0；
Analysis: (1) First of all, we should observe whether the topic has characteristics, and don't blindly do multiplication first. After observation, it is found that the factor on the left side of the equation can be decomposed by the square difference formula and become the product of two linear factors.
(2) The left factor of the equation can be decomposed by cross multiplication.
(3) After it is transformed into a general form, it is solved by the formula method.
(4) Transform the equation into 4x 2-2 (2m+5) x+(m+2) (m+3) = 0, and then decompose it by cross factor.
(1) solution: 4 (x+2) 2-9 (x-3) 2 = 0;
[2(x+2)+3(x-3)][2(x+2)-3(x-3)]= 0；
(5x-5)(-x+ 13)= 0；
5x-5=0 or-x+13 = 0;
∴x 1= 1,x2= 13；
(2) Solution: x2+2x-3 = 0;
[x-(-3)](x- 1)= 0；
X-(-3)=0 or x-1= 0;
∴x 1=-3,x2= 1；
(3) Solution: x2-2x =-;
X 2-2 x+= 0+= 0 (converted to general form first);
△=(-2)^2-4×= 12-8=4>； 0;
∴x=；
∴x 1=,x2=；
(4) Solution: 4x2-4mx-10x+m2+5m+6 = 0;
4x^2-2(2m+5)x+(m+2)(m+3)=0；
[2x-(m+2)][2x-(m+3)]= 0；
2x-(m+2)=0 or 2x-(m+3) = 0;
∴x 1=,x2=；
Example 6. Find two roots of equation 3 (x+1) 2+5 (x+1) (x-4)+2 (x-4) 2 = 0.
Analysis: This equation will be more complicated if it is multiplied first and then merged into a general form. Observing the topic carefully, we find that if x+ 1 and x-4 are regarded as a whole, the left side of the equation can be decomposed by cross multiplication (in fact, method of substitution is used);
Solution: [3 (x+1)+2 (x-4)] [(x+1)+(x-4)] = 0;
That is (5x-5) (2x-3) = 0;
∴5(x- 1)(2x-3)=0；
(x- 1)(2x-3)= 0；
X- 1 = 0 or 2x-3 = 0；;
∴x 1= 1, and x2= is the solution of the original equation.
Example 7. Solving the quadratic equation of one variable about x×2+px+q = 0 by collocation method;
Solution: x 2+px+q = 0 can be transformed into;
X 2+px =-q (constant term moved to the right of the equation);
X 2+px+() 2 =-q+() 2 (the square of both sides of the equation plus half the coefficient of the first term);
(x+)2= (formula);
When P 2-4q ≥0, ≥ 0 (P 2-4q must be discussed separately);
∴x=- =；
∴x 1=,x2=；
When P 2-4q shows that this problem is an equation with letter coefficients, there are no additional conditions for P and Q in the problem, so we should always pay attention to the requirements for letter values in the process of solving the problem and discuss it in categories if necessary.
Exercise:
(1) Solve the following equation with an appropriate method:
1.6x^2-x-2=0 2。 (x+5)(x-5)= 3；
3.x^2-x=0 4.x^2-4x+4=0；
5.3x2+ 1=2x 6。 (2x+3)2+5(2x+3)-6 = 0；
(2) Solve the following equation about x;
1.x^2-ax+-b2=0 2.x^2-(+)ax+a2=0；
Practice reference answer:
( 1) 1.x 1=- 1/2，x2=2/3 2.x 1=2，x2 =-2；
3.x 1=0，x2 = 4 . x 1 = x2 = 2 5 . x 1 = x2 =；
6. Solution: (Take 2x+3 as a whole and decompose the factors on the left side of the equation);
[(2x+3)+6][(2x+3)- 1]= 0；
That is (2x+9) (2x+2) = 0;
* 2x+9 = 0 or 2x+2 = 0；;
∴x 1=-, x2=- 1 is the solution of the original equation.
(2) 1. solution: x 2-ax+(+b) (-b) = 0 2, solution: x 2-(+) ax+aa = 0;
[x-(+b)][x-(-b)]= 0(x-a)(x-a)= 0；
∴x-(+b)=0 or x-(-b)=0 x-a=0 or x-a = 0;
∴x 1=+b, x2=-b is ∴x 1=a, x2=a is;
The solution of the original equation. ;
Multiple choice question:
1. The root of the equation x(x-5)=5(x-5) is ();
a、x=5 B、x=-5 C、x 1=x2=5 D、x 1 = x2 =-5；
2. The value of polynomial a2+4a- 10 is equal to 1 1, so the value of a is ().
A, 3 or 7 B, -3 or 7 C, 3 or -7 D, -3 or-7;
3. If the sum of quadratic coefficient, linear coefficient and constant term in unary quadratic equation AX 2+BX+C = 0 is equal to zero, then the equation must have a root ().
a、0 B、 1 C 、- 1 D、 1；
4. The unary quadratic equation AX 2+BX+C = 0 has a root, if ().
A, b≠0 and c=0 B, b=0 and c ≠ 0;
C, b=0 and c=0 D, c = 0;
5. The two roots of the equation x 2-3x = 10 are ().
a 、-2，5 B、2、-5 C、2，5 D 、-2、-5；
6. The solution of equation x 2-3x+3 = 0 is ().
A, b, c, d, without real roots;
7. The solution of equation 2x 2-0. 15 = 0 is ().
a、x=B、x =-；
c、x 1=0.27，x2=-0.27 D、x 1=，x2 =-；
8. After the left side of the equation x 2-x-4 = 0 is matched into a completely flat mode, the equation obtained is ().
a 、( x-)2=B 、( x-)2 =-；
C, (x-)2=D, none of the above answers are correct;
9. It is known that the unary quadratic equation x 2-2x-m = 0, and the equation after solving the formula of this equation by matching method is ().
a、(x- 1)^2=m2+ 1 B、(x- 1)^2=m- 1 C、(x- 1)^2= 1-m D、(x- 1)^2=m+ 1；
Answer and analysis:
Answer:1.c2.c3.b4.d5.a6.d7.d8.c9.d;
Analysis:
1. Analysis: (x-5) 2 = 0, then x 1=x2=5,
Note: Don't easily divide the two sides of the equation with an algebraic expression. Another quadratic equation with one variable has real roots, and it must be two.
2. analysis: according to the meaning of the question: a 2+4a-10 =1,the solution is a=3 or a=-7.
3. analysis: according to the meaning of the question: if a+b+c=0, the left side of the equation is a+b+c, and only x= 1, ax 2+bx+c = a+b+c, that is, when x= 1.
When the equation is established, there must be a root of x= 1
4. Analysis: the unary quadratic equation AX 2+BX+C = 0 If a root is zero.
Then ax 2+bx+c must have a factor x, and if only c=0, there is a common factor x, so c=0.
In addition, you can also substitute x=0 to get c=0, which is relatively simple!
5. Analysis: The original equation becomes x 2-3x- 10 = 0,
Then (x-5) (x+2) = 0;
X-5=0 or X+2 = 0;
x 1=5，x2 =-2；
6. Analysis: δ = 9-4× 3 =-37. Analysis: 2x2 = 0.15;
x2 =；
x =；
Pay attention to the simplification of roots, square directly and don't lose roots.
8. Analysis: multiply both sides by 3: x 2-3x- 12 = 0, and then according to the linear coefficient formula, x 2-3x+(-) 2 = 12+(-) 2,
The arrangement is as follows: (x-)2= the equation can be deformed by the nature of equality. When x 2-bx is formulated, the term of the formula is the square of half the coefficient of the first term-b.
9. Analysis: x 2-2x = m, then x 2-2x+ 1 = m+ 1, then (x- 1) 2 = m+ 1.